Question: You have found the following ages (in years) of all 4 meerkats at your local zoo: $ 16,\enspace 4,\enspace 16,\enspace 11$ What is the average age of the meerkats at your zoo? What is the variance? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 4 meerkats at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{16 + 4 + 16 + 11}{{4}} = {11.8\text{ years old}} $ Find the squared deviations from the mean for each meerkat. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $16$ years $4.2$ years $17.64$ years $^2$ $4$ years $-7.8$ years $60.84$ years $^2$ $16$ years $4.2$ years $17.64$ years $^2$ $11$ years $-0.8$ years $0.64$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{17.64} + {60.84} + {17.64} + {0.64}} {{4}} $ $ {\sigma^2} = \dfrac{{96.76}}{{4}} = {24.19\text{ years}^2} $ The average meerkat at the zoo is 11.8 years old. The population variance is 24.19 years $^2$.